If $f(x) = e^{-|x|}$, show that $f''(x) - f(x) = -2\delta(x)$ (in the sense of distributions) Exercise :

If $f(x) = e^{-|x|}$, show that for its derivatives, it is : $f''(x)-f(x) = -2\delta(x)$, in the sense of distributions.

Attempt :
I am completely at a loss on how to handle such an exercise as we haven't studied distributions for more than 1-2 lessons, but I know that :
The function $f(x)$ is continuous and differentiable in $\mathbb R \setminus \{0\} = \mathbb R^*$, with derivative :
$$f'(x) = \begin{cases} -e^{-x}, & x>0 \\ e^x, & x<0\end{cases}$$
How would I proceed with modeling the derivatives now in the sense of distributions to prove the equation asked?
To be mentioned, the function $\delta(x)$ is the famous Dirac-delta function, such that:
$$\delta(x) = \begin{cases} +\infty & x=0, \\ 0 & x \neq 0 \end{cases}$$
 A: $\delta$ is the distribution defined by $\langle \varphi,\delta \rangle = \varphi(0)$ for any test function $\varphi$.
Thus we want to show that for any test function $\varphi$,
$$ \langle f''-f,\varphi \rangle = -2\varphi(0). $$
The derivative of a distribution is defined by $\langle f', \varphi \rangle = -\langle f,\varphi' \rangle $, mimicking the integration by parts formula. So we want to show that
$$ \langle f,\varphi''-\varphi \rangle = -2\varphi(0). $$
Everything here is a function, so we can expand the pairing to its definition on functions, namely
$$ \langle f,\varphi''-\varphi \rangle = \int_{-\infty}^{\infty} f(x)(\varphi''(x)-\varphi(x)) \, dx = \int_{-\infty}^0 e^{x} (\varphi''(x)-\varphi(x)) \, dx + \int_0^{\infty} e^{-x} (\varphi''(x)-\varphi(x)) \, dx. $$
Everything here is smooth on the intervals of integration, so we can integrate by parts on the first term in each:
$$ \int_{-\infty}^0 e^{x} (\varphi''(x)-\varphi(x)) \, dx + \int_0^{\infty} e^{-x} (\varphi''(x)-\varphi(x)) \, dx = \left[ e^{x}\varphi'(x) \right]_{-\infty}^0 + \left[ e^{-x}\varphi'(x) \right]_0^{\infty} + \int_{-\infty}^0 e^{x} (-\varphi'(x)-\varphi(x)) \, dx + \int_0^{\infty} e^{-x} (\varphi'(x)-\varphi(x)) \, dx \\
= 0 + \int_{-\infty}^0 e^{x} (-\varphi'(x)-\varphi(x)) \, dx + \int_0^{\infty} e^{-x} (\varphi'(x)-\varphi(x)) \, dx, $$
and then integrating by parts again,
$$ \int_{-\infty}^0 e^{x} (-\varphi'(x)-\varphi(x)) \, dx + \int_0^{\infty} e^{-x} (\varphi'(x)-\varphi(x)) \, dx = [-e^x\varphi(x)]_{-\infty}^0 + [e^{-x}\varphi(x)]_0^{\infty} + \int_{-\infty}^0 e^x(\varphi(x)-\varphi(x)) \, dx + \int_0^{\infty} e^{-x}(+\varphi(x)-\varphi(x)) \, dx \\
= -2\varphi(0), $$
as required.
A: Let $g\in\mathscr S$ be a test function. We have, by definition,
$$
f[g]:=\int_{\mathbb R}g(x)\mathrm e^{-|x|}\mathrm dx\tag1
$$
Also, given $T\in\mathscr S^*$ an arbitrary distribution, we have, by definition,
$$
T'[g]:=-T[g']\tag2
$$
Therefore,
$$
(f''-f)[g]\overset{(2)}\equiv f[g'']-f[g]\overset{(1)}\equiv\int_{\mathbb R}(g''(x)-g(x))\mathrm e^{-|x|}\mathrm dx\tag3
$$
Now integrate by parts:
$$
\begin{aligned}
\int_{\mathbb R^+}(g''(x)-g(x))\mathrm e^{-x}\mathrm dx&\equiv\left[(g'(x)+g(x))\mathrm e^{-x}\right]_{\mathbb R^+}\\
\int_{\mathbb R^-}(g''(x)-g(x))\mathrm e^{+x}\mathrm dx&\equiv\left[(g'(x)-g(x))\mathrm e^{+x}\right]_{\mathbb R^+}
\end{aligned}\tag4
$$
Adding both contributions,
$$
(f''-f)[g]\overset{(4)}\equiv -2g(0)\tag5
$$
But this is precisely $-2\delta[g]$, as required.
--
For a more direct, but slightly less rigorous, approach, you may as well just notice that
$$
f(x)=\mathrm e^{+x}\Theta(-x)+\mathrm e^{-x}\Theta(+x)\tag6
$$
where $\Theta$ is the so-called step-function.
Using $\Theta'(x)=\delta(x)$, you get
$$
f'(x)=\mathrm e^{+x}(\Theta(-x)-\delta(x))+\mathrm e^{-x}(\delta(x)-\Theta(+x))\tag7
$$
and
$$
f''(x)=\mathrm e^{+x}(\Theta(-x)-2\delta(x)-\delta'(x))+\mathrm e^{-x}(-2\delta(x)+\Theta(+x)+\delta'(x))\tag8
$$
Adding both contributions,
$$
f''(x)-f(x)=\mathrm e^{+x}(-2\delta(x)-\delta'(x))+\mathrm e^{-x}(-2\delta(x)+\delta'(x))\tag9
$$
This is precisely $-2\delta(x)$, as required. (Note that $\delta(x)f(x)=\delta(x)f(0)$ and $\delta'(x)f(x)=-\delta(x)f'(0)$).
A: Taking the Fourier transform (with the convention $\hat{f}(\xi)=\int f(x)e^{- ix\xi}dx$) of $f$ we get
$$\hat{f}(\xi)=\frac{2}{1+\xi^2} $$
Hence 
$$\hat{f}+\xi^2\hat{f}=2 $$
The function $f$ can now be identified as a tempered distribution and the above equation still holds in this setting. Taking the inverse Fourier transform of the equation (in the sense of tempered distributions) we get
$$f-f''=2\delta $$
where I used the fact that the (inverse) Fourier transform turns moltiplication by a power into differentiation and that the Fourier transform of $\delta$ is $1$ because for all $f\in \mathcal{S}(\mathbb{R})$ we have 
$$\hat{\delta} (f)=\delta (\hat{f})=\hat{f}(0)=\int 1\cdot f(x)dx=1(f) $$
